The separating words problem asks for the size of the smallest DFA needed to
distinguish between two words of length <= n (by accepting one and rejecting
the other). In this paper we survey what is known and unknown about the
problem, consider some variations, and prove several new results

A. Nozaki

A.W. To

J. Currie

J. Shallit

J.M. Robson

M. Chrobak

P. Goralčík

R.A. Gimadeev

V. Geffert

English

arXiv

Proceedings of the 13th International Workshop, DCFS 2011, Gießen/Limburg, Germany, July 25-27, 2011.The separating words problem asks for the size of the smallest DFA needed to distinguish between two words of length ≤ n (by accepting one and rejecting the other). In this paper we survey what is known and unknown about the problem, consider some variations, and prove several new results

Demaine, Erik D.

Eisenstat, Sarah Charmian

Shallit, Jeffrey

Wilson, David A.

DSpace@MIT

Remarks on Separating WordsErik D. Demaine1, Sarah Eisenstat1, Jeffrey Shallit2, and David A. Wilson11 MIT Computer Science and Artificial Intelligence Laboratory, 32 Vassar Street,Cambridge, MA 02139, USA, {edemaine,seisenst,dwilson}@mit.edu2 University of Waterloo, Waterloo, ON N2L 3G1, Canada,shallit@cs.uwaterloo.caAbstract. The separating words problem asks for the size of the small-est DFA needed to distinguish between two words of length ≤ n (byaccepting one and rejecting the other). In this paper we survey what isknown and unknown about the problem, consider some variations, andprove several new results.1 IntroductionImagine a computing device with very limited powers. What is the simplestcomputational problem you could ask it to solve? It is not the addition of twonumbers, nor sorting, nor string matching — it is telling two inputs apart: dis-tinguishing them in some way.Take as our computational model the deterministic finite automaton or DFA.As usual, it consists of a 5-tuple, M = (Q,Σ, δ, q0, F ), where Q is a finitenonempty set of states, Σ is a nonempty input alphabet, δ : Q × Σ → Q isthe transition function (assumed to be complete, or defined on all members ofits domain), q0 ∈ Q is the initial state, and F ⊆ Q is a set of final states.We say that a DFA M separates w and x if M accepts one but rejects theother. Given two distinct words w, x we let sep(w, x) be the number of statesin the smallest DFA accepting w and rejecting x. For example, the DFA belowseparates 0010 from 1000.0, 1010 1However, by a brief computation, we see that no 2-state DFA can separatethese two words. So sep(1000, 0010) = 3. Note that sep(w, x) = sep(x,w), be-cause the language of a DFA can be complemented by swapping the reject andaccept states.We let S(n) = maxw 6=x|w|,|x|≤nsep(w, x). The separating words problem is to deter-mine good upper and lower bounds on S(n). This problem was introduced 25years ago by Goralcˇ´ık and Koubek [5], who proved S(n) = o(n). It was laterstudied by Robson [7, 8], who obtained the best upper bound so far: S(n) =O(n2/5(log n)3/5).As an additional motivation, the separating words problem can be viewedas an inverse of a classical problem from the early days of automata theory:given two DFAs accepting different languages, what length of word suffices todistinguish them? More precisely, given two DFAs M1 and M2, with m andn states, respectively, with L(M1) 6= L(M2), what is a good bound on thelength of the shortest word accepted by one but not the other? The usual cross-product construction quickly gives an upper bound of mn− 1 (make a DFA forL(M1) ∩ L(M2)). But the optimal upper bound of m + n − 2 follows fromthe usual algorithm for minimizing automata. Furthermore, this bound is bestpossible [9, Thm. 3.10.6]. For NFAs the bound is exponential in m and n [6].From the following result, already proved by Goralcˇ´ık and Koubek [5], weknow that the challenging case of word separation comes from words of equallength:Proposition 1. Suppose |w|, |x| ≤ n and |w| 6= |x|. Then sep(w, x) = O(log n).Furthermore, there is an infinite class of examples where sep(w, x) = Ω(log n).We use the following lemma [10]:Lemma 1. If 0 ≤ i, j ≤ n and i 6= j, then there is a prime p ≤ 4.4 log n suchthat i 6≡ j (mod p).Proof. (of Proposition 1) Let’s prove the upper bound. If |w| 6= |x|, then byLemma 1 there exists a prime p ≤ 4.4 log n such that |w| mod p 6= |x| mod p.Hence a simple cycle of p states serves to distinguish w from x.For the other direction, we first recall that a sequence (pi)i≥0 is said to beultimately periodic if there exist integers r ≥ 0, s ≥ 1 such that pi = pr+i for alli ≥ s. In this case s is called the preperiod and r the period.Now we claim that no DFA with n states can distinguish0n−1 from 0n−1+lcm(1,2,...,n).To see this, let pi = δ(q0, 0i) for i ≥ 0. Then pi is ultimately periodic with period≤ n and preperiod at most n − 1. Thus pn−1 = pn−1+lcm(1,2,...,n). Since, fromthe prime number theorem, we have lcm(1, 2, . . . , n) = en(1+o(1)), the Ω(log n)lower bound follows. ¤Example 1. Suppose |w| = 22 and |x| = 52. Then |w| ≡ 1 (mod 7) and |x| ≡3 (mod 7). So we can accept w and reject x with a DFA that uses a cycle of size7, as follows:0. 10. 10. 1 0. 10. 10. 10. 1In what follows, then, we only consider the case of equal-length words, andwe redefine S(n) = maxw 6=x|w|=|x|=nsep(w, x). The goal of the paper is to survey what isknown and unknown, and to examine some variations on the original problem.Our main new results are Theorems 2 and 3.Notation: in what follows, if x is a word, we let x[j] denote the j’th symbolof x (so that x[1] is the first symbol).2 Independence of alphabet sizeAs we have defined it, S(n) could conceivably depend on the size of the alphabetΣ. Let Sk(n) be the maximum number of states needed to separate two length-nwords over an alphabet of size k. Then we might have a different value Sk(n)depending on k = |Σ|. The following result shows this is not the case for k ≥ 2.This result was stated in [5] without proof; we supply a proof here.Proposition 2. For all k ≥ 2 we have Sk(n) = S2(n).Proof. Suppose x, y are distinct length-n words over an alphabet Σ of size k > 2.Then x and y must differ in some position, say for a 6= b,x = x′ a x′′,y = y′ b y′′,for |x′| = |y′|.Now map a to 0, b to 1 and map all other letters of Σ to 0. This gives twonew distinct binary words X and Y of length n. If X and Y can be separatedby an m-state DFA, then so can x and y, by renaming transitions of the DFAto be over Σ\b and {b} instead of 0 and 1, respectively. Thus Sk(n) ≤ S2(n).But clearly S2(n) ≤ Sk(n), since every binary word can be considered as a wordover the larger alphabet Σ. So Sk(n) = S2(n). ¤3 Average caseOne frustrating aspect of the separating words problem is that nearly all pairs ofwords can be easily separated. This means that bad examples cannot be easilyproduced by random search.Proposition 3. Consider a pair of words (w, x) selected uniformly from the setof all pairs of unequal words of length n over an alphabet of size k. Then theexpected number of states needed to separate w from x is O(1).Proof. With probability 1−1/k, two randomly-chosen words will differ in the firstposition, which can be detected by an automaton with 3 states. With probability(1/k)(1−1/k) the words will agree in the first position, but differ in the second,etc. Hence the expected number of states needed to distinguish two randomly-chosen words is bounded by∑i≥1(i+2)(1/k)i−1(1−1/k) = (3k−2)/(k−1) ≤ 4.¤4 Lower bounds for words of equal lengthFirst of all, there is a lower bound analogous to that in Proposition 1 for wordsof equal length. This does not appear to have been known previously.Theorem 1. No DFA of at most n states can separate the equal-length binarywords w = 0n−11n−1+lcm(1,2,...,n) and x = 0n−1+lcm(1,2,...,n)1n−1.Proof. In pictures, we have0−cycle1−cycle0−tail1−tailMore formally, let M be any DFA with n states, let q be any state, and leta be any letter. Let pi = δ(q, ai) for i ≥ 0. Then pi is ultimately periodic withperiod ≤ n and preperiod (“tail”) at most n−1. Thus pn−1 = pn−1+lcm(1,2,...,n).It follows that after processing 0n−1 and 0n−1+lcm(1,2,...,n), M must be in thesame state. Similarly, after processing0n−11n−1+lcm(1,2,...,n) and 0n−1+lcm(1,2,...,n)1n−1,M must be in the same state. So no n-state machine can separate w from x. ¤We now prove a series of very simple results showing that if w and x differin some “easy-to-detect” way, then sep(w, x) is small.4.1 Differences near the beginning or end of wordsProposition 4. Suppose w and x are words that differ in some symbol thatoccurs d positions from the start. Then sep(w, x) ≤ d+ 2.Proof. Let t be a prefix of length d of w. Then t is not a prefix of x. We canaccept the language tΣ∗ using d + 2 states; such an automaton accepts w andrejects x. ¤For example, to separate01010011101100110000from01001111101011100101we can build a DFA to recognize words that begin with 0101:0, 11010(Transitions to a dead state are omitted.)Proposition 5. Suppose w and x differ in some symbol that occurs d positionsfrom the end. Then sep(w, x) ≤ d+ 1.Proof. Let the DFA M be the usual pattern-recognizing automaton for thelength-d suffix s of w, ending in an accepting state if the suffix is recognized.Then M accepts w but rejects x. States of M correspond to prefixes of s, andδ(t, a) = the longest suffix of ta that is a prefix of s. ¤For example, to separate11111010011001010101from11111011010010101101we can build a DFA to recognize those words that end in 0101:0011 0110104.2 FingerprintsDefine |w|a as the number of occurrences of the symbol a in the word w.Proposition 6. If |w|, |x| ≤ n and |w|a 6= |x|a for some symbol a, then sep(w, x) =O(log n).Proof. By the prime number theorem, if |w|, |x| = n, and w and x have k and moccurrences of a respectively (k 6= m), then there is a prime p = O(log n) suchthat k 6≡ m (mod p). So we can separate w from x just by counting the numberof a’s, modulo p. ¤Analogously, we have the following result.Proposition 7. If there is a pattern of length d occurring a different number oftimes in w and x, with |w|, |x| ≤ n, then sep(w, x) = O(d log n).4.3 Pairs with low Hamming distanceThe previous results have shown that if w and x have differing “fingerprints”,then they are easy to separate. By contrast, the next result shows that if w andx are very similar, then they are also easy to separate.The Hamming distance H(w, x) between two equal-length words w and x isdefined to be the number of positions where they differ.Theorem 2. Let w and x be words of length n. If H(w, x) ≤ d, then sep(w, x) =O(d log n).Proof. Without loss of generality, assume x and y are binary words, and x hasa 1 in some position where y has a 0. Consider the following picture:1 nidi4i3i2i1y =x = · · ·· · ·01Let i1 < i2 < . . . < id be the positions where x and y differ. Now considerN = (i2−i1)(i3−i1) · · · (id−i1). Then N < nd−1. By the prime number theorem,there exists some prime p = O(logN) = O(d log n) such that N is not divisibleby p. So ij 6≡ i1 (mod p) for 2 ≤ j ≤ d.Define ap,k(x) = ∑j≡k (mod p)x[j] mod 2. This value can be calculated bya DFA consisting of two connected rings of p states each. We use such a DFAcalculating ap,i1 . Since p is not a factor of N , none of the positions i2, i3, . . . , idare included in the count ap,i1 , and the two words x and y agree in all otherpositions. So x contains exactly one more 1 in these positions than y does, andhence we can separate the two words using O(d log n) states. ¤5 Special classes of words5.1 ReversalsIt is natural to think that pairs of words that are related might be easier toseparate than arbitrary words; for example, it might be easy to separate a wordfrom its reversal. No better upper bound is known for this special case. However,we still have a lower bound of Ω(log n) for this restricted problem:Proposition 8. There exists a class of words w for which sep(w,wR) = Ω(log n)where n = |w|.Proof. Consider separatingw = 0t−110t−1+lcm(1,2,...t)fromwR = 0t−1+lcm(1,2,...t)10t−1.Then, as before, no DFA with ≤ t states can separate w from wR. ¤Must sep(wR, xR) = sep(w, x)? No, for w = 1000, x = 0010, we havesep(w, x) = 3butsep(wR, xR) = 2.Open Problem 1 Is∣∣sep(x,w)− sep(xR, wR)∣∣ unbounded?5.2 ConjugatesTwo words w,w′ are conjugates if one is a cyclic shift of the other. For example,the English words enlist and listen are conjugates. Is the separating wordsproblem any easier if restricted to pairs of conjugates?Proposition 9. There exist a infinite class of pairs of words w, x such that w, xare conjugates, and sep(w, x) = Ω(log n) for |w| = |x| = n.Proof. Consider againw = 0t−110t−1+lcm(1,2,...t)1andw′ = 0t−1+lcm(1,2,...t)10t−11.¤6 Nondeterministic separationWe can define nsep(w, x) in analogy with sep: the number of states in the smallestNFA accepting w but rejecting x. There do not seem to be any published resultsabout this measure.Now there is an asymmetry in the inputs: nsep(w, x) need not equal nsep(x,w).For example, the following 2-state NFA accepts w = 000100 and rejects x =010000, so nsep(w, x) ≤ 2.00, 10However, an easy computation shows that there is no 2-state NFA acceptingx and rejecting w, so nsep(x,w) ≥ 3.Open Problem 2 Is |nsep(x,w)− nsep(w, x)| unbounded?A natural question is whether NFAs give more separation power than DFAs.Indeed they do, since sep(0001, 0111) = 3 but nsep(0001, 0111) = 2. However,a more interesting question is the extent to which nondeterminism helps withseparation — for example, whether it contributes only a constant factor or thereis any asymptotic improvement in the number of states required.Theorem 3. The quantity sep(w, x)/nsep(w, x) is unbounded.Proof. Consider once again the wordsw = 0t−1+lcm(1,2,...,t)1t−1 and x = 0t−11t−1+lcm(1,2,...,t)where t = n2 − 3n+ 2, n ≥ 4.We know from Theorem 1 that any DFA separating these words must haveat least t+ 1 = n2 − 3n+ 3 states.Now consider the following NFA M :0 0 00 000 0 0 0 001loop of n− 1 statesloop of n statesThe language accepted by this NFA is {0a : a ∈ A}1∗, where A is the setof all integers representable by a non-negative integer linear combination of nand n − 1. But t − 1 = n2 − 3n + 1 6∈ A, as can be seen by computing t − 1modulo n−1 and modulo n. On the other hand, every integer ≥ t is in A. Hencew = 0t−1+lcm(1,2,...,t)1t−1 is accepted by M but x = 0t−11t−1+lcm(1,2,...,t) is not.NowM has 2n = Θ(√t) states, so sep(x,w)/nsep(x,w) = Ω(√t) = Ω(√log |x|),which is unbounded. ¤Open Problem 3 Find better bounds on sep(w, x)/nsep(w, x).We can also get an Ω(log n) lower bound for nondeterministic separation.Theorem 4. No NFA of n states can separate0n2−11n2−1+lcm(1,2,...,n)from0n2−1+lcm(1,2,...,n)1n2−1.Proof. A result of Chrobak [1], as corrected by To [11], states that every unaryn-state NFA is equivalent to one consisting of a “tail” of at most O(n2) states,followed by a single nondeterministic state that leads to a set of cycles, each ofwhich has at most n states. The size of the tail was proved to be at most n2− 2by Geffert [3].Now we use the same argument as for DFAs above. ¤Open Problem 4 Find better bounds on nsep(w, x) for |w| = |x| = n, as afunction of n.Theorem 5. We have nsep(w, x) = nsep(wR, xR).Proof. Let M = (Q,Σ, δ, q0, F ) be an NFA with the smallest number of statesaccepting w and rejecting x. Now create a new NFA M ′ with initial state equalto any single state in δ(q0, w) ∩ F and final state q0, and all other transitionsof M reversed. Then M ′ accepts wR. But M ′ rejects xR. For if M ′ accepted xRthen M would also accept x, since the input string and transitions are reversed.¤7 Separation by 2DPDA’sIn [2], the authors showed that words can be separated with small context-freegrammars (and hence small PDA’s). In this section we observeProposition 10. Two distinct words of length n can be separated by a 2DPDAof size O(log n).Proof. Recall that a 2DPDA is a deterministic pushdown automaton, with end-markers surrounding the input, and two-way access to the input tape. Given dis-tinct strings w, x of length n, they must differ in some position p with 1 ≤ p ≤ n.Using O(log p) states, we can reach position p on the input tape and accept if(say) the corresponding character equals w[p], and reject otherwise.Here is how to access position p of the input. We show how to go fromscanning position i to position 2i using a constant number of states: we moveleft on the input, pushing two symbols per move on the stack, until the leftendmarker is reached. Now we move right, popping one symbol per move, untilthe initial stack symbol is reached. Using this as a subroutine, and applying itto the binary expansion of p, we can, using O(log p) states, reach position p ofthe input. ¤8 Permutation automataWe conclude by relating the separating words problem to a natural problem ofalgebra.Instead of arbitrary automata, we could restrict our attention to automatawhere each letter induces a permutation of the states (“permutation automata”),as suggested by Robson [8]. He obtained an O(n1/2) upper bound in this case.For an n-state automaton, the action of each letter can be viewed as anelement of Sn, the symmetric group on n elements.Turning the problem around, then, we could ask: what is the shortest pairof distinct equal-length binary words w, x, such that for all morphisms σ :{0, 1}∗ → Sn we have σ(w) = σ(x)? Although one might suspect that the answeris lcm(1, 2, . . . , n), for n = 4, there is a shorter pair (of length 11): 00000011011and 11011000000.Now if σ(w) = σ(x) for all σ, then (if we define σ(x−1) = σ(x)−1) we havethat σ(wx−1) = the identity permutation for all σ.Call any nonempty word y over the letters 0, 1, 0−1, 1−1 an identical relationif σ(y) = the identity for all morphisms σ. We say y is nontrivial if y containsno occurrences of 00−1 and 11−1.What is the length ` of the shortest nontrivial identical relation over Sn?Recently Gimadeev and Vyalyi [4] proved ` = 2O(√n log n).9 AcknowledgmentsThanks to Martin Demaine and Anna Lubiw, who participated in the openproblem session at MIT where these problems were discussed.We thank the DCFS 2011 referees for their helpful suggestions.References1. M. Chrobak. Finite automata and unary languages. Theoret. Comput. Sci., 47:149–158, 1986. Erratum in 302:497–498, 2003.2. J. Currie, H. Petersen, J. M. Robson, and J. Shallit. Separating words with smallgrammars. J. Automata, Languages, and Combinatorics, 4:101–110, 1999.3. V. Geffert. Magic numbers in the state hierarchy of finite automata. Inform.Comput., 205:1652–1670, 2007.4. R. A. Gimadeev and M. N. Vyalyi. Identical relations in symmetric groups andseparating words with reversible automata. In F. M. Ablayev and E. W. Mayr,editors, 5th Int’l Computer Science Symposium in Russia, CSR 2010, volume 6072of Lecture Notes in Computer Science, pages 144–155. Springer-Verlag, 2010.5. P. Goralcˇ´ık and V. Koubek. On discerning words by automata. In L. Kott, editor,Proc. 13th Int’l Conf. on Automata, Languages, and Programming (ICALP), vol-ume 226 of Lecture Notes in Computer Science, pages 116–122. Springer-Verlag,1986.6. A. Nozaki. Equivalence problem of non-deterministic finite automata. J. Comput.System Sci., 18:8–17, 1979.7. J. M. Robson. Separating strings with small automata. Inform. Process. Lett.,30:209–214, 1989.8. J. M. Robson. Separating words with machines and groups. RAIRO Inform. The´or.App., 30:81–86, 1996.9. J. Shallit. A Second Course in Formal Languages and Automata Theory. Cam-bridge University Press, 2009.10. J. Shallit and Y. Breitbart. Automaticity I: Properties of a measure of descriptionalcomplexity. J. Comput. System Sci., 53:10–25, 1996.11. A. W. To. Unary finite automata vs. arithmetic progressions. Inform. Process.Lett., 109:1010–1014, 2009.

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